OpenStudy (anonymous):

anyone have a good way of doing flux integrals?

5 years ago
OpenStudy (anonymous):

does anyone have any idea what i'm taking about?

5 years ago
OpenStudy (anonymous):

Yes. Your question is a bit vague, though. Care to be more specific?

5 years ago
OpenStudy (anonymous):

not really, i just have a hard time with them, and my exam is tmr, so if you have any tricks or methods that are simple, they would be greatly appreciated. i'm goin to get a question

5 years ago
OpenStudy (anonymous):

I don't know if this site has any hints, but it would be the best place to find tricks and tips. http://www.tricki.org/

5 years ago
OpenStudy (anonymous):

Post an example that you're having trouble with and I'll try to work through it with you?

5 years ago
OpenStudy (anonymous):

$\int\limits_{}^{}\int\limits_{}^{} F*ds \rightarrow F=yj -zk$ with the surfaces given by $y=x^2+z^2, 0\le y \le 1, and , x^2+z^2\le1$ with positive orientation

5 years ago
OpenStudy (anonymous):

Okay, so let your paramaters for your surface be x and z. So we have the parametrization: $\Phi(x,z)= \left\{ x=x,z=z,y=x^2+z^2 \right\}$ So we have the flux is: $\iint_V \vec{F} \cdot dS = \iint \vec{F}(\Phi) \cdot ( \frac{\partial \Phi}{\partial x} \times \frac{\partial \Phi}{\partial z})dxdz$ If its over a vector field. Is it a vector field or scalar?

5 years ago
OpenStudy (anonymous):

yea its a vector field

5 years ago
OpenStudy (anonymous):

Wait...lol...let me get a piece of paper

5 years ago
OpenStudy (anonymous):

I'm getting the flux to be zero x.x

5 years ago
OpenStudy (anonymous):

thats the answer, how did you get that?

5 years ago
OpenStudy (anonymous):

i'm completely serious too

5 years ago
OpenStudy (anonymous):

Really? O.O

5 years ago
OpenStudy (anonymous):

I can type out an exposition if you want one lol :P

5 years ago
OpenStudy (anonymous):

don't bother with the equation tool, i've can read it plain text, but if you could at least do the set-up of the integrsl, that would be nice

5 years ago
OpenStudy (anonymous):

I'll type a selection: Like I said use the parameterization: $\vec{\Phi}=(x,x^2+z^2,z)$ From this you find: $\vec{\Phi}_x=(1,2x,0); \vec{\Phi}_z=(0,2z,1)$ From that cross product I get: $(2x,-1,2z) \implies \vec{F}(\vec{\Phi}) \cdot (2x,-1,2z) \implies \iint (0,x^2+z^2,-z) \cdot (2x,-1,2z) dx dz$ To find the bounds use: x^2+z^2 <=1

5 years ago
OpenStudy (anonymous):

ok ive only been using this ite for like 2 hours now, how do give you a medal?

5 years ago
OpenStudy (anonymous):

$\implies -\sqrt{1-z^2} \le x \le \sqrt{1-z^2} \implies -1 \le z \le 1$ $\int\limits_{-1}^1 \int\limits_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} (-(x^2+z^2)-2z^2) dx dz$ And that integral is zero I believe.

5 years ago
OpenStudy (anonymous):

thanks a lot, that actually made a lot of sense

5 years ago
OpenStudy (anonymous):

Well, I did something in my cross product or something...because the integral is -pi...But other than that it should be the right approach

5 years ago
OpenStudy (anonymous):

I don't spot the problem either. :\ Sorry, my math skills are decaying. I need to review this stuff.

5 years ago
OpenStudy (anonymous):

ok, well you forgot that the disk a y=1 is also a surface and i found that to be pi, so when you add those togeter, the total answer is zero

5 years ago
OpenStudy (anonymous):

there was no problem in ur cross porduct

5 years ago
OpenStudy (anonymous):

I did it on paper and got zero again. I'm not sure what wolfram was doing. Once you get that integral set up convert to polar Then I get: $\int\limits_0^{2 \pi} \int\limits_0^1 (-\rho^3-2\rho^3\sin^2(\omega))d \rho d \omega$ Solving that I get: $\frac{- \pi}{2}+\frac{\pi}{2} +\frac{1}{8}\sin(2 \omega)|_0^{2 \pi}=0$ With the help of some trig identities and polar as I said.

5 years ago
OpenStudy (anonymous):

Remember:$dA=dxdz=\rho d \rho d \omega$

5 years ago
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